How to get a value from a cell of a dataframe?

If you have a DataFrame with only one row, then access the first (only) row as a Series using iloc, and then the value using the column name:

In [3]: sub_df
Out[3]:
          A         B
2 -0.133653 -0.030854

In [4]: sub_df.iloc[0]
Out[4]:
A   -0.133653
B   -0.030854
Name: 2, dtype: float64

In [5]: sub_df.iloc[0]['A']
Out[5]: -0.13365288513107493

These are fast access for scalars

In [15]: df = pandas.DataFrame(numpy.random.randn(5,3),columns=list('ABC'))

In [16]: df
Out[16]: 
          A         B         C
0 -0.074172 -0.090626  0.038272
1 -0.128545  0.762088 -0.714816
2  0.201498 -0.734963  0.558397
3  1.563307 -1.186415  0.848246
4  0.205171  0.962514  0.037709

In [17]: df.iat[0,0]
Out[17]: -0.074171888537611502

In [18]: df.at[0,'A']
Out[18]: -0.074171888537611502

You can turn your 1x1 dataframe into a numpy array, then access the first and only value of that array:

val = d2['col_name'].values[0]

Most answers are using iloc which is good for selection by position.

If you need selection-by-label loc would be more convenient.

For getting a value explicitly (equiv to deprecated df.get_value('a','A')) # this is also equivalent to df1.at['a','A'] In [55]: df1.loc['a', 'A'] Out[55]: 0.13200317033032932

It doesn't need to be complicated:

val = df.loc[df.wd==1, 'col_name'].values[0]

I needed the value of one cell, selected by column and index names. This solution worked for me:

original_conversion_frequency.loc[1,:].values[0]

It looks like changes after pandas 10.1/13.1

I upgraded from 10.1 to 13.1, before iloc is not available.

Now with 13.1, iloc[0]['label'] gets a single value array rather than a scalar.

Like this:

lastprice=stock.iloc[-1]['Close']

Output:

date
2014-02-26 118.2
name:Close, dtype: float64

The quickest/easiest options I have found are the following. 501 represents the row index.

df.at[501,'column_name']
df.get_value(501,'column_name')

Not sure if this is a good practice, but I noticed I can also get just the value by casting the series as float.

e.g.

rate

3 0.042679 Name: Unemployment_rate, dtype: float64

float(rate)

0.0426789

df_gdp.columns

Index([u'Country', u'Country Code', u'Indicator Name', u'Indicator Code', u'1960', u'1961', u'1962', u'1963', u'1964', u'1965', u'1966', u'1967', u'1968', u'1969', u'1970', u'1971', u'1972', u'1973', u'1974', u'1975', u'1976', u'1977', u'1978', u'1979', u'1980', u'1981', u'1982', u'1983', u'1984', u'1985', u'1986', u'1987', u'1988', u'1989', u'1990', u'1991', u'1992', u'1993', u'1994', u'1995', u'1996', u'1997', u'1998', u'1999', u'2000', u'2001', u'2002', u'2003', u'2004', u'2005', u'2006', u'2007', u'2008', u'2009', u'2010', u'2011', u'2012', u'2013', u'2014', u'2015', u'2016'], dtype='object')

df_gdp[df_gdp["Country Code"] == "USA"]["1996"].values[0]

8100000000000.0

For pandas 0.10, where iloc is unavalable, filter a DF and get the first row data for the column VALUE:

df_filt = df[df['C1'] == C1val & df['C2'] == C2val]
result = df_filt.get_value(df_filt.index[0],'VALUE')

if there is more then 1 row filtered, obtain the first row value. There will be an exception if the filter result in empty data frame.

This is quite old by now but as of today you can fix it by simply doing

val = float(d2['col_name'].iloc[0])

Converting it to integer worked for me:

int(sub_df.iloc[0])

I've run across this when using DataFrames with MultiIndexes and found squeeze useful.

From the docs:

Squeeze 1 dimensional axis objects into scalars. Series or DataFrames with a single element are squeezed to a scalar. DataFrames with a single column or a single row are squeezed to a Series. Otherwise the object is unchanged.

# example for DataFrame with MultiIndex
> import pandas as pd

> df = pd.DataFrame(
                    [
                        [1, 2, 3], 
                        [4, 5, 6], 
                        [7, 8, 9]
                    ], 
                    index=pd.MultiIndex.from_tuples( [('i', 1), ('ii', 2), ('iii', 3)] ),
                    columns=pd.MultiIndex.from_tuples( [('A', 'a'), ('B', 'b'), ('C', 'c')] )
)

> df
       A  B  C
       a  b  c
i   1  1  2  3
ii  2  4  5  6
iii 3  7  8  9

> df.loc['ii', 'B']
   b
2  5

> df.loc['ii', 'B'].squeeze()
5

Note that while df.at[] also works (if you aren't needing to use conditionals) you then still AFAIK need to specify all levels of the MultiIndex.

Example:

> df.at[('ii', 2), ('B', 'b')]
5

I have a DataFrame with a 6-level index and 2-level columns, so only having to specify the outer level is quite helpful.

Using .item() returns a scalar (not a Series), and it only works if there is a single element selected. It's much safer than .values[0] which will return the first element regardless of how many are selected.

>>> df = pd.DataFrame({'a': [1,2,2], 'b': [4,5,6]})
>>> df[df['a'] == 1]['a']  # Returns a Series
0    1
Name: a, dtype: int64
>>> df[df['a'] == 1]['a'].item()
1
>>> df2 = df[df['a'] == 2]
>>> df2['b']
1    5
2    6
Name: b, dtype: int64
>>> df2['b'].values[0]
5
>>> df2['b'].item()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python3/dist-packages/pandas/core/base.py", line 331, in item
    raise ValueError("can only convert an array of size 1 to a Python scalar")
ValueError: can only convert an array of size 1 to a Python scalar

To get the full row's value as JSON (instead of a Serie):

row = df.iloc[0]

Use the to_json method like bellow:

row.to_json()